0

Let $E$ be a real Banach space. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$ and $\mathcal K(E)$ its subspace consisting of compact operators. For $T \in \mathcal L(E)$,

  • we denote by $N(T)$ its kernel and by $R(T)$ its range.
  • we denote by $\rho(T)$ its resolvent set, by $\sigma(T)$ its spectrum, and by $EV(T)$ its set of eigenvalues. Then $EV(T) \subset \sigma(T) = \mathbb R \setminus \rho(T)$.
  • for $\lambda \in EV(T)$, the set $N(T-\lambda I)$ is called the eigenspace corresponding to $\lambda$.

I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,

Let $E := \ell^2$ with its norm $|\cdot|_2$. An element $x\in E$ is denoted by $x=(x_1, x_2, \ldots, x_n,\ldots)$. Consider the operators $$ \begin{align*} S_r x &= (0, x_1, x_2, \ldots, x_{n-1}, \ldots), \\ S_{\ell} x &= (x_2, x_3, x_4, \ldots, x_{n+1}, \ldots), \end{align*} $$ respectively called the right shift and left shift.

  1. Determine $\|S_r\|$ and $\|S_{\ell}\|$. Does $S_r$ or $S_{\ell}$ belong to $\mathcal{K}(E)$?
  2. Prove that $EV(S_r) = \emptyset$.
  3. Prove that $\sigma\left(S_r\right)=[-1,1]$.
  4. Prove that $E V (S_{\ell}) = (-1, 1)$. Determine the corresponding eigenspaces.
  5. Prove that $\sigma(S_{\ell})=[-1,1]$.
  6. Determine the adjoints $S_r^*$ or $S_{\ell}^*$.
  7. Prove that for every $\lambda \in(-1,1)$, the spaces $R (S_r-\lambda I)$ and $R (S_{\ell}-\lambda I)$ are closed. Give an explicit representation of these spaces.
  8. Prove that the spaces $R(S_r \pm I)$ and $R(S_{\ell} \pm I)$ are dense and that they are not closed.

Consider the multiplication operator $M$ defined by $$ M x = (\alpha_1 x_1, \alpha_2 x_2, \ldots, \alpha_n x_n, \ldots), $$ where $(\alpha_n)$ is a bounded sequence in $\mathbb{R}$.

  1. Determine $E V (S_r \circ M)$.

There are possibly subtle mistakes that I could not recognize in below attempt of (9). Could you please have a check on it? I'm also happy to see other approaches.

We write $S_r M := S_r \circ M$. Then $$ S_r M x = (0, \alpha_1 x_1, \alpha_2 x_2, \ldots, \alpha_n x_n, \ldots). $$

Let $\lambda \in \mathbb R$. Then $$ (S_r M - \lambda I) x = (-\lambda x_1, \alpha_1 x_1 - \lambda x_2, \alpha_2 x_2 - \lambda x_3, \ldots, \alpha_n x_n - \lambda x_{n+1}, \ldots). $$

Then $x \in N (S_r M - \lambda I)$ iff $$ 0 = -\lambda x_1 = \alpha_n x_n - \lambda x_{n+1} \quad \forall n \ge 1. $$

Let $x \in N (S_r M - \lambda I)$.

  • If $\lambda \neq 0$, then $x_n = 0$ for $n \ge 1$. Then $EV (S_r M) = \emptyset$.
  • Assume that $\lambda=0$. Then $\alpha_n x_n =0$ for all $n \ge 1$. If there is $m \ge 1$ such that $\alpha_m = 0$, then $x_m \in \mathbb R$ and thus $EV (S_r M) = \{0\}$. If $\alpha_n \neq 0$ for all $n \ge 1$, then $x_n=0$ for $n \ge 1$ and thus $EV (S_r M) = \emptyset$.
Akira
  • 17,367

0 Answers0